本文目錄一覽:
Opencv問題, 怎樣去除二值圖像中面積較小的連通域?
//=======函數實現=====================================================================
void RemoveSmallRegion(Mat Src, Mat Dst, int AreaLimit, int CheckMode, int NeihborMode)
{
int RemoveCount = 0;
//新建一幅標籤圖像初始化為0像素點,為了記錄每個像素點檢驗狀態的標籤,0代表未檢查,1代表正在檢查,2代表檢查不合格(需要反轉顏色),3代表檢查合格或不需檢查
//初始化的圖像全部為0,未檢查
Mat PointLabel = Mat::zeros(Src.size(), CV_8UC1);
if (CheckMode == 1)//去除小連通區域的白色點
{
//cout “去除小連通域.”;
for (int i = 0; i Src.rows; i++)
{
for (int j = 0; j Src.cols; j++)
{
if (Src.atuchar(i, j) 10)
{
PointLabel.atuchar(i, j) = 3;//將背景黑色點標記為合格,像素為3
}
}
}
}
else//去除孔洞,黑色點像素
{
//cout “去除孔洞”;
for (int i = 0; i Src.rows; i++)
{
for (int j = 0; j Src.cols; j++)
{
if (Src.atuchar(i, j) 10)
{
PointLabel.atuchar(i, j) = 3;//如果原圖是白色區域,標記為合格,像素為3
}
}
}
}
vectorPoint2iNeihborPos;//將鄰域壓進容器
NeihborPos.push_back(Point2i(-1, 0));
NeihborPos.push_back(Point2i(1, 0));
NeihborPos.push_back(Point2i(0, -1));
NeihborPos.push_back(Point2i(0, 1));
if (NeihborMode == 1)
{
//cout “Neighbor mode: 8鄰域.” endl;
NeihborPos.push_back(Point2i(-1, -1));
NeihborPos.push_back(Point2i(-1, 1));
NeihborPos.push_back(Point2i(1, -1));
NeihborPos.push_back(Point2i(1, 1));
}
else int a = 0;//cout “Neighbor mode: 4鄰域.” endl;
int NeihborCount = 4 + 4 * NeihborMode;
int CurrX = 0, CurrY = 0;
//開始檢測
for (int i = 0; i Src.rows; i++)
{
for (int j = 0; j Src.cols; j++)
{
if (PointLabel.atuchar(i, j) == 0)//標籤圖像像素點為0,表示還未檢查的不合格點
{ //開始檢查
vectorPoint2iGrowBuffer;//記錄檢查像素點的個數
GrowBuffer.push_back(Point2i(j, i));
PointLabel.atuchar(i, j) = 1;//標記為正在檢查
int CheckResult = 0;
for (int z = 0; z GrowBuffer.size(); z++)
{
for (int q = 0; q NeihborCount; q++)
{
CurrX = GrowBuffer.at(z).x + NeihborPos.at(q).x;
CurrY = GrowBuffer.at(z).y + NeihborPos.at(q).y;
if (CurrX = 0 CurrXSrc.colsCurrY = 0 CurrYSrc.rows) //防止越界
{
if (PointLabel.atuchar(CurrY, CurrX) == 0)
{
GrowBuffer.push_back(Point2i(CurrX, CurrY)); //鄰域點加入buffer
PointLabel.atuchar(CurrY, CurrX) = 1; //更新鄰域點的檢查標籤,避免重複檢查
}
}
}
}
if (GrowBuffer.size()AreaLimit) //判斷結果(是否超出限定的大小),1為未超出,2為超出
CheckResult = 2;
else
{
CheckResult = 1;
RemoveCount++;//記錄有多少區域被去除
}
for (int z = 0; z GrowBuffer.size(); z++)
{
CurrX = GrowBuffer.at(z).x;
CurrY = GrowBuffer.at(z).y;
PointLabel.atuchar(CurrY, CurrX) += CheckResult;//標記不合格的像素點,像素值為2
}
//********結束該點處的檢查**********
}
}
}
CheckMode = 255 * (1 – CheckMode);
//開始反轉面積過小的區域
for (int i = 0; i Src.rows; ++i)
{
for (int j = 0; j Src.cols; ++j)
{
if (PointLabel.atuchar(i, j) == 2)
{
Dst.atuchar(i, j) = CheckMode;
}
else if (PointLabel.atuchar(i, j) == 3)
{
Dst.atuchar(i, j) = Src.atuchar(i, j);
}
}
}
//cout RemoveCount ” objects removed.” endl;
}
//=======函數實現=====================================================================
//=======調用函數=====================================================================
Mat img;
img = imread(“D:\\1_1.jpg”, 0);//讀取圖片
threshold(img, img, 128, 255, CV_THRESH_BINARY_INV);
imshow(“去除前”, img);
Mat img1;
RemoveSmallRegion(img, img, 200, 0, 1);
imshow(“去除後”, img);
waitKey(0);
//=======調用函數=====================================================================
python matplotlib subplot 上面面積大下面小怎麼辦
在matplotlib下,一個Figure對象可以包含多個子圖(Axes),可以使用subplot()快速繪製,其調用形式如下:
subplot(numRows, numCols, plotNum)
圖表的整個繪圖區域被分成numRows行和numCols列,plotNum參數指定創建的Axes對象所在的區域,如何理解呢?
如果numRows = 3,numCols = 2,那整個繪製圖表樣式為3X2的圖片區域,用坐標表示為(1,1),(1,2),(1,3),(2,1),(2,2),(2,3)。這時,當plotNum = 1時,表示的坐標為(1,3),即第一行第一列的子圖;
import numpy as np
python如何刪除二值化圖片中小塊白色區域
如果確定是純白的話你就把rgb都小於某個極小常數的像素點的alpha設成0就好了你說的nodata應該就是透明的意思!
原創文章,作者:小藍,如若轉載,請註明出處:https://www.506064.com/zh-hk/n/309336.html